Let $h(x)=6 x^5-15 x^4+10 x^3$. On which intervals is $h$ increasing? Choose 1 answer: Choose 1 answer: (Choice A) A $(1, \infty)$ only (Choice B) B $(-\infty,1)$ only (Choice C) C $(-\infty,0)$ and $(1, \infty)$ (Choice D) D $(0,1)$ only (Choice E) E The entire domain of $h$
Explanation: We can analyze the intervals where $h$ is increasing/decreasing by looking for the intervals where its derivative $h'$ is positive/negative. A function can only change its direction from increasing to decreasing and vice versa between its critical points and the points where the function itself is undefined. The derivative of $h$ is $h'(x)=30x^2(x-1)^2$. $h'(x)=0$ for $x=0,1$. Since $h'$ is a polynomial, it's defined for all real numbers. Therefore, our critical points are $x=0$ and $x=1$. $h$ is defined for all real numbers so we only need to consider the critical points. Our critical points divide the number line into three intervals: $\llap{-}1$ $0$ $1$ $2$ $(-\infty,0)$ $(0,1 )$ $(1,\infty)$ Let's evaluate $h'$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $h'(x)$ Verdict $(-\infty,0)$ $x=-1$ $h'(-1)=120>0$ $h$ is increasing $\nearrow$ $(0,1)$ $x=\dfrac{1}{2}$ $h'\left(\dfrac12\right)=\dfrac{15}{8}>0$ $h$ is increasing $\nearrow$ $(1,\infty)$ $x=2$ $h'(2)=120>0$ $h$ is increasing $\nearrow$ In conclusion, $h$ is increasing over its entire domain (which is all real numbers).